Equivalence of definitions of Frattini subgroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term Frattini subgroup
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

Definition in terms of maximal subgroups

For a group $G$, the Frattini subgroup $\Phi (G)$ is defined as the intersection of all maximal subgroups of $G$:

$\Phi (G)=\bigcap _{M\leq _{max}G}M$

When there are no maximal subgroups, we define $\Phi (G)$ as the whole group $G$

Definition as the set of nongenerators

For a group $G$, the Frattini subgroup $\Phi (G)$ is defined as the set of all nongenerators. An element $x\in G$ is termed a nongenerator if, whenever $S\subset G$ is such that $S\cup \{x\}$ generates $G$, $S$ itself generates $G$.

Facts used

• Every 1-completed subgroup is contained in a maximal subgroup

Proof

The set of nongenerators is contained in every maximal subgroup

Given: A group $G$, a nongenerator $x$ for $G$, a maximal subgroup $M$ of $G$

To prove: $x\in M$

Proof: Suppose (contradiction point) $x\notin M$. Then, $\langle M,x\rangle =G$. Thus, the set $M\cup \{x\}$ is a generating set for $G$. By the definition of nongenerator, we see that $M$, as a set, is a generating set for $G$. But this is a contradiction since the subgroup generated by $M$ is $M$ itself.

Thus, $x\in M$ is forced.

Any element in the intersection of all maximal subgroups is a nongenerator

Given: A group $G$, an element $x$ in the intersection of maximal subgroups of $G$, a subset $S$ such that $S\cup \{x\}$ generates $G$

To prove: $S$ generates $G$

Proof: Let $H$ be the subgroup generated by $S$. Suppose (contradiction point) $H$ is proper. Then, $x\notin H$, since $\langle H,x\rangle =G$. Thus, $H$ is a 1-completed subgroup: it, along with one outside element, generates the whole group. By the above fact, there exists a maximal subgroup $M$ containing $H$. But by assumption, $x$ is in every maximal subgroup, so $x\in M$, so $\langle H,x\rangle \leq M$, so $G\leq M$, a contradiction.

Thus, the assumption that $H$ is proper is false, so $H=G$, so $S$ generates $G$.

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Exercise 25, Page 199 (Section 6.2)